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Indian Classical LiteratureChapter Unit

Module – 1 Multiple Integrals

Topics Covered:

  • Evaluation of double integration in Cartesian and plane polar coordinates
  • Evaluation of double integration by changing the order of integration
  • Area as a double integral (Cartesian)
  • Area as a double integral (Polar)
  • Triple integration in Cartesian coordinates
  • Conversion from Cartesian to polar in double integrals
  • Volume using triple integrals
  • Applications of multiple integrals in Engineering

Evaluation of double integration – Cartesian and Polar coordinates

Type – 1 Limits are constants

  1. Evaluate ∫01∫02(x2+y2) dx dy.\int_{0}^{1} \int_{0}^{2} (x^2 + y^2) \, dx \, dy.

    Solution:

    ∫01∫02(x2+y2) dx dy=∫01[x33+x y2]x=0x=2 dy\int_{0}^{1} \int_{0}^{2} (x^2 + y^2) \, dx \, dy = \int_{0}^{1} \left[\frac{x^3}{3} + x \, y^2 \right]_{x=0}^{x=2} \, dy =∫01[(83+2y2)−(13+y2)] dy= \int_{0}^{1} \left[\left(\frac{8}{3} + 2y^2\right) - \left(\frac{1}{3} + y^2\right)\right] \, dy =∫01(73+y2) dy= \int_{0}^{1} \left(\frac{7}{3} + y^2\right) \, dy =[73y+y33]y=0y=1= \left[\frac{7}{3}y + \frac{y^3}{3}\right]_{y=0}^{y=1} =73+13=83.= \frac{7}{3} + \frac{1}{3} = \frac{8}{3}.

    Note: ∫01∫02(x2+y2) dy dx=83.\int_{0}^{1} \int_{0}^{2} (x^2 + y^2) \, dy \, dx = \frac{8}{3}.

    Observation: If the limits of integration are constants, then the order of integration is insignificant.


  1. Evaluate ∫03∫02x y(x+y) dy dx.\int_{0}^{3} \int_{0}^{2} x \, y (x + y) \, dy \, dx.

    Solution:

    ∫03∫02x y(x+y) dy dx=∫03∫02(x2 y+x y2) dy dx\int_{0}^{3} \int_{0}^{2} x \, y (x + y) \, dy \, dx = \int_{0}^{3} \int_{0}^{2} (x^2 \, y + x \, y^2) \, dy \, dx =∫03[x2 y22+x y33]y=0y=2 dx= \int_{0}^{3} \left[\frac{x^2 \, y^2}{2} + \frac{x \, y^3}{3}\right]_{y=0}^{y=2} \, dx =∫03[x2⋅42+x⋅83] dx= \int_{0}^{3} \left[\frac{x^2 \cdot 4}{2} + \frac{x \cdot 8}{3}\right] \, dx =∫03(2x2+8x3) dx= \int_{0}^{3} \left(2x^2 + \frac{8x}{3}\right) \, dx =[2x33+8x26]x=0x=3= \left[\frac{2x^3}{3} + \frac{8x^2}{6}\right]_{x=0}^{x=3} =[2⋅273+8⋅96]= \left[\frac{2 \cdot 27}{3} + \frac{8 \cdot 9}{6}\right] =18+12=30.= 18 + 12 = 30.

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